- 1. The slides used in this video are color coded. If you are
experiencing difficulty with one aspect of your understanding than
another you might find this coding useful!Slides with Slides with
tan red backgrounds backgrounds involve involve word matching
Slides with green problems. concepts. backgrounds Slides with olive
involve backgrounds graphing. involve reading data tables.
2. Dr. Fiala is traveling on his Harley at a constant 13.67 m/s.
What is the distance traveled by Doc in 7.32 seconds? 3. SOLUTION:
Find the distance traveled.Vi&f = 13.67 m/s Xf ti = 0 s tf =
7.32 s Xi = 0 m Xf = m a = 0 m/s2 4. SOLUTION: Find the distance
traveled.Vi&f = 13.67 m/s Xf ti = 0 s tf = 7.32 s Xi = 0 m Xf=
100.07 m a = 0 m/s2 5. Dr. Fiala notices he is now traveling at a
constant 49.21 km/h. What is the distance in meters traveled by Doc
in 7.32 seconds? 6. SOLUTION: Dimensional analysis. M K H D
1xxX0dcxXX49.21 km/h = 13.67 m/s So the Xf remains 100.07 mm 7. Dr.
Fiala jumps in his unstarted car. He accelerates at a rate of 4
m/s2 for 8 seconds. How far did Doc travel? 8. Docs final position.
vi = 0 m/s ti = 0 s tf = 8 s a = 4 m/s2 Xi = 0 mvf = Xf =Xf = 128 m
9. Displacement Velocity Acceleration Inertia Force Momentum The
change in the rate or direction of motion. The resistance to a
change in an objects current state of motion. A change in position.
A push or a pull that tends to accelerate an object. The movement
of an object in a specific direction over time. The product of mass
times velocity. 10. Displacement is a change in position. Velocity
is the movement of an object in a specific direction over time.
Acceleration is the change in the rate or direction of motion of an
object. Inertia is the resistance to a change in an objects current
state of motion. Force is a push or a pull that tends to accelerate
an object. Momentum is the product of mass times velocity. 11. Time
(s) 1Object #1 Object #2 Object #3 Object #4 Position (m) Position
(m) Position (m) Position (m) 1642344482416326 12. Time (s)Object
#1 Object #2 Object #3 Object #4 Position (m) Position (m) Position
(m) Position (m)11648223281643481224646416328 13. Time (s) 1Object
#1 Object #2 Object #3 Object #4 Velocity (m/s) Velocity (m/s)
Velocity (m/s) Velocity (m/s) 168 92 3 4138 102 4 14. Time
(s)Object #1 Object #2 Object #3 Object #4 Velocity (m/s) Velocity
(m/s) Velocity (m/s) Velocity (m/s)116482214.56943138106411.510118
15. #1#3#2#4 16. 2m /sm/s 10 m/s0 m/s 17. Vy = 0 m/sYf = Yi + Vi t
+ gt2 Vi = 48.5 m/s ty = 5 sVf2 = Vi2 + 2g y Vi = 48.5 m/s 18.
Graph Options Position GraphCan you predict the slope shape and
orientation of both the velocity and acceleration graphs? 19. Graph
Options Position GraphCan you predict the slope shape and
orientation of both the velocity and acceleration graphs?V = 0 m/s
20. Graph Options Position GraphCan you predict the slope shape and
orientation of both the velocity and acceleration graphs?V = 0 m/s
21. Graph Options Position GraphCan you predict the slope shape and
orientation of both the velocity and acceleration graphs? 22.
Position Graph Vy = 0 m/sVelocity Graph Acceleration Graph+V Vy = 0
m/s -Vg = -9.8 m/s2 23. If Dr. Fiala starts from a full stop and
accelerates at 2.25 m/s2, how incredibly fast will he be traveling
when he has traveled 1530 meters? 24. SOLUTION: Calculate final
velocity without knowing time.Vi = 0 m/s Xi = 0 m Xf = 1530 m a =
2.25 m/s2 ti = 0 sVf tfVf= 82.98 m/s 25. If Dr. Fiala starts from a
full stop and accelerates at 2.25 m/s2, how long will it take him
to drive 1530 meters? 26. SOLUTION: Solve for time.Vi = 0 m/s Xi =
0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf = 82.98 m/stf Vft = 36.88
s 27. 0 m/s2 2.530 m2m/s 2.93m/s 750 m/s233.75 m 15 m90 m 28.
Velocity(m/s) 16 14 12 10 8 6 4 2 002468101214 16 Time (s) 29.
Velocity(m/s) 16 14 12 10 8 6 4 2 002468101214 16 Time (s) 30.
Graph Options Motion Map Can you predict the slope shape and
orientation of the position, velocity, and acceleration graphs? 31.
Graph Options Motion Map Can you predict the slope shape and
orientation of the position, velocity, and acceleration graphs? 32.
Graph Options Motion Map Can you predict the slope shape and
orientation of the position, velocity, and acceleration graphs? 33.
Graph Options Motion Map Can you predict the slope shape and
orientation of the position, velocity, and acceleration graphs? 34.
Position GraphMotion MapVelocity GraphAcceleration Graph 35. Graph
Options Motion MapCan you predict the slope shape and orientation
of the position, velocity, and acceleration graphs? 36. Graph
Options Motion MapCan you predict the slope shape and orientation
of the position, velocity, and acceleration graphs? 37. Graph
Options Motion MapVelocity GraphPosition GraphCan you predict the
slope shape and orientation of the position, velocity, and
acceleration graphs? 38. Graph Options Motion MapVelocity
GraphPosition GraphCan you predict the slope shape and orientation
of the position, velocity, and acceleration graphs? 39. Position
GraphMotion MapVelocity GraphAcceleration Graph 40. Graph Options
Motion MapCan you predict the slope shape and orientation of the
position, velocity, and acceleration graphs? 41. Graph Options
Motion MapCan you predict the slope shape and orientation of the
position, velocity, and acceleration graphs?Position Graph 42.
Graph Options Motion MapCan you predict the slope shape and
orientation of the position, velocity, and acceleration
graphs?Position Graph 43. Graph Options Motion Map Velocity
GraphCan you predict the slope shape and orientation of the
position, velocity, and acceleration graphs?Position Graph 44.
Position GraphMotion MapVelocity GraphAcceleration Graph 45. 0
m/s318 m/s 46. 0 m/s3 -118 m/sm/ s317.5 m/s0 m/s3 2 m/s 47. 2.25 +
3.25 = 2.25 + 3.25 = 2.25 + 3.25 = 48. SOLUTION: Adding vectors.
+2.25 + 3.25 = 5.5+2.25 + 3.25 = 1.00+2.25 + 3.25 = 3.95+-++-+ 49.
? kg153 NDetermine the mass of a 153.08 N object. 50. SOLUTION:
Calculate mass.W = 153.08 N g = -9.8 m/s2mm = 15.62 kg 51.
Determine the force of friction on a 15.62 kg object traveling at a
constant horizontal velocity of 3.62 m/s while experiencing an
applied force of 6 N. 52. SOLUTION: Calculate force of friction.m =
51.62 kg a = 0 m/s2 g = -9.8 m/s2 Fa = 6 NFfFf = -6 N 53. Force
Diagrams Motion MapMatch the force diagram to the motion map. Can
you also predict the slope shape and orientation of the position,
velocity, and acceleration graphs? 54. Force Diagrams Motion
MapMatch the force diagram to the motion map. Can you also predict
the slope shape and orientation of the position, velocity, and
acceleration graphs? 55. Force Diagram Motion Map Position
GraphVelocity GraphAcceleration Graph 56. Determine the force
needed to accelerate Dr. Fialas car and its occupants at a rate of
3.23 m/s2 if the total mass of car and occupants is 1315 kg and
there is no friction force. 57. SOLUTION: Find applied force.m =
1315 kg a = 3.23 m/s2FF = 4247.45 N (kg)(m/s ) 2 58. This time,
when we apply that 4247.45 N force to Dr. Fialas car and its
occupants, the resulting acceleration is actually lower. It
registers at a rate of only 3.00 m/s2. What is the magnitude for
the force of friction causing the acceleration to be decreased? 59.
SOLUTION: Find applied force.m = 1315 kg F = 4247.45 N a = 3.00
m/s2FfFf = - 302.45 N 60. Force Diagrams Motion MapMatch the force
diagram to the motion map. Can you also predict the slope shape and
orientation of the position, velocity, and acceleration graphs? 61.
Force Diagrams Motion MapMatch the force diagram to the motion map.
Can you also predict the slope shape and orientation of the
position, velocity, and acceleration graphs? 62. Force Diagram
Motion Map Position GraphVelocity GraphAcceleration Graph 63. When
an object is freefalling it is weightless. Prove mathematically
that a .448 kg apple is weightless during its freefall from a tree.
Draw a force diagram of the apple during its fall from the tree.
64. SOLUTION: Find force of support.m = .448 kg g = -9.8 m/s2FsFs =
0 N Fg = -4.39 N 65. Motion Map Force Diagram Position (Y)
GraphVelocity (Vy) Graph Can you predict the motion map, and
kinematic graphs for this freefalling object?Acceleration Graph 66.
Motion MapForce DiagramPosition (Y) GraphVelocity (Vy)
GraphAcceleration Graph 67. Assuming a perfectly frictionless
surface, ideal for launching students in a game of faculty bowling,
Dr. Fiala uses a brand new gizmo that automatically applies a force
that results in an acceleration of 1.1 m/s2. Experimentation
resulted in a student with a mass of 44.10 kg, accelerating at 1.1
m/s2. Find the force generated by the gizmo for that student. 68.
SOLUTION: Find force in the horizontal.m = 44.10 kg a = 1.1 m/s2FF
= 48.51 N 69. Mass (kg)Force (N)004246.2 47.564448.4 48.65 49.25
49.5145.4550 70. Mass (kg)Force
(N)004246.243.2547.564448.444.2348.6544.7749.2545.0149.5145.4550
71. All of the students from the previous problem (combined mass)
step into an elevator at the same time. Draw a force diagram of
this situation including the magnitude of Fg and Fs. 72. SOLUTION:
Find force of gravity and force of support. m1 = 42 kg Fg m2 =
43.25 kg Fs m3 = 44 kg m4 = 44.23 kg m5 = 44.77 kg m6 = 45.01 kg m2
= 45.45 kg Fg= -3025.36 N g = -9.8 m/s2 Fs= 3025.36 N 73. This same
elevator accelerates at a rate of .75 m/s2 towards the second
floor. Draw a force diagram of this situation including the
magnitude of Fg and Fs. 74. SOLUTION: Find force of support. m =
308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .75 m/s2FsFs = 3256.89
N Fs=N Fg = 3025.36 N3256.89 75. Motion Map Force Diagram Position
GraphVelocity (Vy) Graph Can you predict the motion map, and
kinematic graphs for this elevator?Acceleration Graph 76. Motion
MapForce DiagramPosition GraphVelocity (Vy) GraphAcceleration Graph
77. This same elevator accelerates 2 at a rate of .50 m/s as it
begins its stop for the second floor. Draw a Force diagram of this
situation including the magnitude of Fg and Fs. 78. SOLUTION: Find
force of support. m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a =
.-50 m/s2 NFs = 2871.01 N Fg = 3025.36 NFsFs= 2871.01 79. Motion
Map Force Diagram Position GraphVelocity (Vy) Graph Can you predict
the motion map, and kinematic graphs for the ENTIRE
TRIP?Acceleration Graph 80. Motion MapForce DiagramPosition
GraphVelocity (Vy) GraphAcceleration Graph 81. According to Newtons
3rd law, an action force causes an equal on opposite reaction
force. It is no wonder a truck windshield squashes a bug and not
vice versa. A 2000 kg truck and a .0002 kg bug hit with a 50 N
force. Take a closer look at why the truck wins the collision by
calculating the acceleration exerienced by the bug and by the
truck. 82. SOLUTION: Why the bug doesnt survive.mt = 2000 kg at mb
= .0002 kg ab g = -9.8 m/s2 F = -50 N at = -.025 m/s2 ab = -250,000
m/s2 83. .40 N 375These cables will snap if the mass of the
trafffic light exceeds 10.1 kg. Does the traffic light exceed 10.1
kg? 84. SOLUTION: The cable does not break.T1 = 375.4 N g = -9.8
m/s2 = 7.5m T1y m= 10 kg 85. Dr. Fiala attempts to walk due east at
5 m/s at the same time as a 30 m/s cold, winter wind is blowing due
south. What is the magnitude of Dr. Fialas velocity. 86. SOLUTION:
Resultant velocity magnitude.Vi = 30 m/s Vi = 5 m/s Vx = 5 m/sVy =
30 m/sVfVf= 30.41 m/s22a +b =c2 87. Vx = 5 m/s V=30.4 1m /sVy = 30
m/sIf Dr. Fiala continues his velocity and the wind continues to
blow steadily, at what angle, as measured from positive X, is Dr.
Fialas velocity. 88. SOLUTION: Resultant velocity angle measured
from positive x. tan = x y = 9.46Vx = 5 m/sVy = 30 m/stan = y x =
80.54 (from +x) = 279.46 89. Because of this wind, a 15 kg package
is blown from Dr. Fialas arms and onto the ground. The 15 kg
package reaches a velocity of 30.41 m/s in a time of 4 seconds.
Find the force acting on the box horizontally if there is no
friction. 90. SOLUTION: Find applied force. Yf = -15 m a Yi = 0 m F
m = 15 kg g = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/s ti = 0 s a = 7.85
m/s2 ti = 4 s F = 117.79 N 91. Force Diagrams Motion Map Match the
force diagram to the motion map. Can you also predict the slope
shape and orientation of the position, velocity, and acceleration
graphs? 92. Force Diagram Motion Map Position GraphVelocity
GraphAcceleration Graph 93. Force Ff = 0 N Diagram Motion MapFs =
147 N Fa = 117.79 N Fg = 147 NPosition GraphVelocity Graph 31.4
m/s62.8 m 4sAcceleration Graph 7.85 m/s24s4s 94. If the package is
blow horizontally at 30.41 m/s off a ledge onto a parking lot that
is 15 meters below how much time will it spend in the air before
striking the ground? What does the motion map look like? 95.
SOLUTION: Find time package spends in the air.Yf = -15 m Yi = 0 m m
= 15 kg g = -9.8 m/s2 Vi = 0 m/s ti = 0 m/sVf tf tf = 1.75 s 96.
Force Diagram Motion Map Position (Y) GraphVelocity (Vy) Graph Can
you predict what the force diagram, and vertical kinematic graphs
for this freefalling object?Acceleration Graph 97. Force Diagram
Motion MapPosition (Y) GraphVelocity (Vy) GraphAcceleration Graph
98. Time (s)Vertical Position (m)Horizontal Position (m)Vertical
Velocity Horizontal Velocity (m/s) (m/s)0.11 0.2731.41
-0.360.63-5.61-4.80-1.941.07-2.6515.390.498.481.22-6.17 33.61
38.321.35-8.931.46-10.441.75--11.96 -13.2345.86 99. Time
(s)Vertical Position (m)Horizontal Position (m)Vertical Velocity
Horizontal Velocity (m/s)
(m/s)0.11-0.063.46-1.0831.410.27-0.368.48-2.6531.410.49-1.1815.39-4.8031.410.63-1.9419.79-6.1731.411.07-5.6133.61-10.4931.411.22-7.2938.32-11.9631.411.35-8.9342.40-13.2331.411.46-10.4445.86-14.3131.411.75-15.0154.97-17.1531.41
100. Dr. Fiala throws a baseball in the air with an initial
velocity of 27 m/s at an angle of 27 to the horizon. Create a
velocity vector diagram and show, by parallelogram method, the X
and Y components of the baseballs velocity. 101. SOLUTION:Resolve
velocity vector into x and y components just like force or any
other vector.V = 27 m/s = 27 g = -9.8 m/s2 m/s 27 27Viy Vix Viy =
12.26 m/s Vix = 24.06 m/sVx = V Vy = V 102. How much time will it
take for the baseball to reach the same height from which it was
thrown? 103. SOLUTION: Find time in the air. g = -9.8 m/s2 = 27 Viy
= 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 mtftf = 2.5 s
104. How far will the baseball travel in 2.5 seconds? 105.
SOLUTION: Find range. g = -9.8 m/s2 Xf = 27 Viy = 12.26 m/s Viy =
24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m tf = 2.5 s Xf = 60.15
m 106. What is the maximum height the baseball attained during its
flight? 107. SOLUTION: Find y. g = -9.8 m/s2 y = 27 Viy = 12.26 m/s
Vix = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m y = 7.67 m tf =
2.5 s 108. Vy = 0 m/sa=9.8 mVf = Vi + 2g t tf = 3.06 s-/s 2Yf = Yi
+ Vi t + at2 Yf = 45.9 m AREA = Base x Height 109. Force Vector
Arrows for this ProjectilePositionVelocityUsing these vector arrows
can you predict what the position, force, velocity and acceleration
vector arrows would look like for this projectile at the start and
at the top?Acceleration 110. Force Position Velocity Acceleration
111. If it was a .448 kg apple that was thrown into the air at 30
m/s what was the apples intial momentum? 112. SOLUTION: Find
momentum of apple.m = .448 kg Vi = 30 m/s g = -9.8 m/s2pp = 13.44
kgm/s 113. What constant force is needed to get a change in the
apples momentum from 13.44 kgm/s to 0 In 3.06 seconds? 114.
SOLUTION: Find force necessary to change momentum.m = .448 kg F Vi
= 30 m/s g = -9.8 m/s2 ti = 0 s tf = 3.06 s p = -13.44 kgm/s F =
4.39 N 115. After falling to the ground the .448 kg apple rolled at
a constant 10.4 m/s where collided with a stationary .577 kg apple.
If the two apples stuck together, at what velocity would they roll?
116. SOLUTION: Find the velocity of two apples stuck together.m1 =
.448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/sVf pp
= 4.66 kgm/s2 Vf = 4.55 m/s 117. Determine the force applied if the
rolling apples strike a wall and a come to a stop in .311 seconds.
118. SOLUTION: Find force needed to stop apples. m1 = .448 kg m2 =
.577 kg ti = 0 s tf = .311 s g = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0
m/s p = 4.66 kgm/sFF = 14.98 N
